Study Tools

The Taylor Series

Problems

Problems

It is instructive to compute the Taylor series for several of the elementary functions. We
take up this task in the present section.

The Sine and Cosine Functions

We compute the first few derivatives of
f (x) = sin(x)
:

f'(x)

= cos(x)

f(2)(x)

= - sin(x)

f(3)(x)

= - cos(x)

f(4)(x)

= sin(x)

At this point, we have arrived back at
sin(x)
, and the pattern will repeat itself.
Evaluating the derivatives at
0
, we find that

f (0)

= 0

f'(0)

= 1

f(2)(0)

= 0

f(3)(0)

= - 1

f(4)(0)

= 0

and so on. This allows us to write down the first few terms of the Taylor series for
f (x) = sin(x)
at
0
:

x - + - + ...

Below, we plot the graph of
sin(x)
, together with the graphs of its Taylor polynomials
p1(x)
,
p3(x)
, and
p5(x)
.

Figure %: Plot of
sin(x)
,
p1(x)
,
p3(x)
, and
p5(x)

We can write the entire Taylor series as

x2n+1

This series has infinite radius of convergence. Noting that
| f(n)(x)|≤1
for all
n≥ 0
(since
f(n)(x)
is equal to
sin(x)
,
cos(x)
,
-sin(x)
, or
-cos(x)
), we get a bound on the remainder term:

| rn(x)| = xn≤| x|n

Since
= 0
for all real numbers
x
, we have
| rn(x)| = 0
, so the Taylor series for
f (x) = sin(x)
converges to
f (x)
for all
x
.

The situation with
g(x) = cos(x)
is similar. The first few derivatives at
0
are

g(0)

= 1

g'(0)

= 0

g(2)(0)

= - 1

g(3)(0)

= 0

g(4)(0)

= 1

so the Taylor series at
0
looks like

1 - + - + ... = x2n

Again the radius of convergence is infinite, and the same estimate applies to the remainder term
as for the sine function, so
cos(x)
coincides with the function represented by its Taylor
series for all
x
.

The Exponential Function

The exponential function has perhaps the simplest Taylor series of all. If
f (x) = ex
,
then
f(n)(x) = ex
for all
n≥ 0
, so the Taylor series for
f
at
0
is just

1 + x + + + ... =

since
e0 = 1
. For any given
x
, the remainder term satisfies the inequality

| rn(x)| = ec = ec

for some
c
between
0
and
x
. Since
c≤x
implies
ec≤ex
, a constant,
and
= 0
, we once again have that
| rn(x)| = 0
, so the function defined by the Taylor series for
f (x) = ex
is equal to
f (x)
for all
x
.

The Logarithm Function

This time, since
log(0)
is not defined, we choose instead to find the Taylor series for
f (x) = log(x)
at
x = 1
. We compute the first few derivatives below:

f'(x)

=

f(2)(x)

= -

f(3)(x)

=

f(2)(x)

= -

Thus

f(n)(x) = (- 1)n-1

for
n≥1
, so

f(n)(1) = (- 1)n-1(n - 1)!

and the Taylor series for
f (x) = log(x)
at
x = 1
is

(x - 1) - + - + ... = (- 1)n-1

In this case, it turns out that the series only converges (to the
value
log(x)
) for
x
in the interval
(0, 2)
, i.e. the radius of
convergence is
1
. Letting
y = x - 1
, we may write